ta có : \(A=\dfrac{sin^2\alpha-cos^2\alpha}{sin\alpha.cos\alpha}=\dfrac{\dfrac{sin^2\alpha}{cos^2\alpha}-\dfrac{cos^2\alpha}{cos^2\alpha}}{\dfrac{sin\alpha.cos\alpha}{cos^2\alpha}}\) \(=\dfrac{tan^2\alpha-1}{tan\alpha}\)
\(\Leftrightarrow A=\dfrac{\left(\sqrt{3}\right)^2-1}{\sqrt{2}}=\sqrt{2}\) vậy \(A=\sqrt{2}\)